How many Cl2O molecules of different masses (isotopologues) naturally exist due to the presence of stable isotopes of chlorine and oxygen?

• A. 9
• B. 3
• C. 6
• D. 2

Answer: (A)

Verified step by step guidance

1. Identify the stable isotopes of chlorine and oxygen that contribute to different isotopologues. Chlorine has two stable isotopes: \(^{35}\mathrm{Cl}\) and \(^{37}\mathrm{Cl}\). Oxygen has three stable isotopes: \(^{16}\mathrm{O}\), \(^{17}\mathrm{O}\), and \(^{18}\mathrm{O}\).
2. Determine the number of possible isotope combinations for the chlorine atoms in \(\mathrm{Cl}_2\). Since there are two chlorine atoms, each can be either \(^{35}\mathrm{Cl}\) or \(^{37}\mathrm{Cl}\), so the number of chlorine isotope combinations is \(2 \times 2 = 4\).
3. Determine the number of possible isotope combinations for the oxygen atom in \(\mathrm{O}\). Since there is only one oxygen atom, the number of oxygen isotope combinations is equal to the number of stable oxygen isotopes, which is 3.
4. Calculate the total number of isotopologues by multiplying the number of chlorine isotope combinations by the number of oxygen isotope combinations: \(4 \times 3 = 12\).
5. Consider that some isotopologues may be indistinguishable due to symmetry or identical masses. For example, the two chlorine atoms are identical, so combinations like \(^{35}\mathrm{Cl}-^{37}\mathrm{Cl}\) and \(^{37}\mathrm{Cl}-^{35}\mathrm{Cl}\) represent the same isotopologue. Adjust the count accordingly by using combinations with repetition for chlorine isotopes: the number of unique chlorine isotope combinations is \(\frac{2(2+1)}{2} = 3\). Then multiply by 3 oxygen isotopes to get \(3 \times 3 = 9\) unique isotopologues.